Fix an imaginary quadratic field $K$, and let $\mathcal{O}_K$ be its ring of integers. A Hermitian modular form of genus 1 (i.e., an automorphic form on $GU(1,1)$) of weight $(k_1,k_2)$ on a congruence subgroup $\Gamma \subset U(1,1)(\mathbb{Z})$ is a function $f \colon \mathfrak{h} \to \mathbb{C}$ satisfying an analytic condition and a transformation law:

$$ f\left(\frac{az+b}{cz+d}\right) = (cz+d)^{k_1}(\overline{c}z+\overline{d})^{k_2} f(z), \qquad \forall \begin{pmatrix} a & b \\ c & d\end{pmatrix} \in \Gamma. $$

Here $\Gamma \subset GU(1,1)(\mathbb{Z})$ is a group of matrices with entries in $\mathcal{O}_K$, so conjugating the coefficients makes sense. At this stage these are very similar to classical modular forms, and in fact they have the same shape $q$-expansions and simply restricting the matrices that are allowed to act on them allows them to be viewed as classical modular forms of weight $k_1+k_2$ on the congruence subgroup $\Gamma \cap SL_2(\mathbb{Z})$.

I have not found anywhere what the (complex) $L$-functions for these things are. For a classical modular cusp form, we have $$ f(z) = \sum_{n=1}^\infty a_n q^n, \qquad \Rightarrow \qquad L(f,s) = \sum_{n=1}^\infty \frac{a_n}{n^s}. $$

Is there a similar formula for what the $L$-function of a hermitian cusp form of genus 1 should be? Is it the same, maybe modulo the Euler factors at infinity? How are the $L$-functions of the hermitian modular form of weight $(k_1,k_2)$ and the classical modular form of weight $k_1+k_2$ related?

Is there a similar relationship between the hermitian modular forms over a CM field $F$ and the Hilbert modular forms over its totally real subfield $F^+$?

Follow up: how should I view the fact that every Hermitian modular form yields a classical modular form, while an elliptic curve over $\mathbb{Q}$ yields an elliptic curve over $K$? If I equate "elliptic curve over $\mathbb{Q}$" to "classical modular form on $\Gamma_0(N)$," and "elliptic curve over $K$" to "hermitian modular form on a similar $\Gamma$," these feel like they go in opposite directions from each other.


1 Answer 1


To define an "L-function" for an automorphic representation $\pi$ on a group $G$ you usually have to make a choice of a representation $\rho$ of the dual group. Only then can you define $L(\pi,\rho,s)$. If $G = \mathrm{GL}(n)$ then there is an obvious choice of representation so it is often ommitted in that case. In the cases you are looking at, the most natural $L$-function will coincide on the nose with the $L$-function of the modular form.

Yes, the situation over totally real fields is very similar.

Followup: First, you need to be more precise. An elliptic curve over $\mathbf{Q}$ should correspond to a modular form on $\Gamma_0(N)$ of weight two and with rational coefficients.

Second, (and this is the real problem), an elliptic curve over $K$ defintely won't in general give rise to an Hermetian modular form, it should give rise to a form on the (quite different) group $\mathrm{GL}(2)/K$. The motives you do get from Hermetian modular forms have the property that they are conjugate self dual (up to a specific range of twists). Since $2$-dimensional representations are always self dual up to twist anyway (since $\mathrm{GL}(2) = \mathrm{GSp}(2)$), this means the corresponding motives are both conjugate self-dual and self-dual up to twist, which means that (if they are not induced) they are isomorphic to their conjugates up to twist. As it happens, this basically implies that the representations themselves up to twist descend to $\mathbf{Q}$, and so it turns out they have to come from $\mathrm{GL}(2)$ anyway. In particular, you get both fewer motives and fewer autormophic forms for $U(1,1)$.

One final wrinkle: you might say: given an elliptic curve $E/\mathbf{Q}$, the restriction $E/K$ certainly has an isomorphic Tate module to its conjugate. But why is that the only thing that can happen? After all, we are only requiring that the Tate module of $E/K$ is isomorphic to a twist of the Tate module of $E^c/K$. This could happen, for example, if $E$ is isogenous to $E^c$ over some finite extension $L/K$. (The Tate conjecture, proved in this case by Faltings, means that this is the only way this can happen.) Well, you would be correct, and indeed these elliptic curves do exist! In this case, one calls $E$ a $\mathbf{Q}$-curve. The $\mathbf{Q}$-curves do give rise to Hermetian modular forms even though they need not come from $\mathbf{Q}$. But this is not inconsistent with what I said previously, it's just that the modular forms corresponding to these $\mathbf{Q}$-curves - while still of weight $2$ - don't have rational coefficients (and they might have level $\Gamma_1(N)$ as well rather than $\Gamma_0(N)$).

  • $\begingroup$ Is the representation of the dual group you're alluding to like the spin representation in the case of Siegel modular forms? (As opposed to the standard representation?) Or is it different enough that the analogy there isn't useful? And for the follow-up, thank you for the insight! I thought we looked at U(1,1) instead of GL(2) because it was somehow necessary to get the similar picture with elliptic curves. $\endgroup$
    – Jon Aycock
    Jun 26, 2019 at 21:02
  • $\begingroup$ It is exactly an example of this. If $G = \mathrm{GSp}(2n)$, then the dual group is $\mathrm{Gspin}(2n+1)$. The latter group has many representations, including a "spin" representation of dimension $2^n$ and a "standard" representation of dimension $2n+1$, so specifying that representation is part of the data defining an $L$-function for $\mathrm{GSp}(2n)$. $\endgroup$ Jun 26, 2019 at 23:20
  • $\begingroup$ For $n = 2$, these degree $4$ and $5$ $L$-functions are the classical ones which are known to have analytic continuations. (Note $\mathrm{Gspin}(5) \simeq \mathrm{GSp}(4)$ by some accidental automorphism.) But there are other $L$-functions, for example, $\mathrm{GSp}(4)$ has a (unique) irreducible representation $\rho$ of dimension $91$, so there should also be an $L$-function $L(\pi,\rho,s)$ for a Siegel modular form. (It exists, but its analytic continuation is not known.) $\endgroup$ Jun 26, 2019 at 23:25
  • $\begingroup$ How far can this U(n,n) <--> GSp(2n) analogy go? Does the Hecke algebra look pretty much the same for U(n,n) as it does for GSp(2n)? In particular, similar looking Satake parameters, with spin and standard L-functions for U(n,n) having Euler products looking the same as those for GSp(2n)? (I see a lot more information on Siegel than Hermitian modular forms, and would like to know what I can use!) $\endgroup$
    – Jon Aycock
    Jun 27, 2019 at 9:07
  • $\begingroup$ Your comment is pretty vague. But you should really just learn the recipe of how all these things connect (Satake parameters, Langlands dual group, L-functions)are connected and then the conjectural story is just Lie theory. For example, the dual group of $U(n,n)$ is something close to $GL(2n)$ semi-direct product with $\mathrm{Gal}(K/\mathbf{\mathbf{Q}})$ acting by conjugate transpose. So $L$-functions related to representations of this group. Similarly, Satake paremeters are related to elements of this group. (When $n = 1$ this automorphism is inner, which is why things are simpler.) $\endgroup$ Jun 27, 2019 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.